Ram was asked to calculate the sum of an arithmetic progression whose ...
Let us calculate the actual sum which Ram would have got if he did not have had made the mistake.
We can use the formula to calculate the sum = 30/2 (2(2) + (30 − 1)7)
Thus the actual sum = 15(4 + 29 × 7) = 3105
Excess sum calculated = 3210 - 3105 = 105
Let last n terms be the terms where the common difference was taken as 14 instead of 7
Thus 7 + 14 + ...n terms = 105
n/2(2(7) + (n − 1)7) = 105
n(n + 1)7 = 210
Or 2 + n − 30 = 0 or (n + 6)(n − 5) = 0.
Since n can not be negative, n = 5, Thus last 5 terms had incorrect common differece. x = 30 - n = 30 - 5 = 25
x − 4 = 21 = 3 × 7
Total 4 factors are there.
Ram was asked to calculate the sum of an arithmetic progression whose ...
To find the sum of an arithmetic progression, we can use the formula:
S = (n/2)(2a + (n-1)d)
where S is the sum, n is the number of terms, a is the first term, and d is the common difference.
In this case, the sum of the arithmetic progression with a first term of 2 and a common difference of 7 is:
S = (30/2)(2(2) + (30-1)(7))
= 15(4 + 203)
= 15(207)
= 3105
So, the sum of the first 30 terms of the arithmetic progression with a common difference of 7 is 3105.
Now, let's consider the sum of the first x terms of the arithmetic progression with a common difference of 14. We'll use the same formula, but with a new common difference:
S' = (x/2)(2(2) + (x-1)(14))
We know that S' = 3210, so we can solve for x:
3210 = (x/2)(4 + 14x - 14)
3210 = (x/2)(14x - 10)
6420 = x(14x - 10)
6420 = 14x^2 - 10x
14x^2 - 10x - 6420 = 0
By factoring or using the quadratic formula, we find that the two possible values for x are x = 30 or x = -38. Since we can't have a negative number of terms, we discard x = -38.
Therefore, the number of factors of x is the number of factors of 30, which is 8: 1, 2, 3, 5, 6, 10, 15, 30.